我不确定我应该在这里或在数学网站上询问,但我会在这里试一试。下面是作业:用户输入10个数字,即域 D 。我需要评估声明:
For All x, y in D, x<y or y <2x
的。我的问题是什么是x,什么是y?
如果我输入1 2 3 4 5 6 7 8 9 10,因为域x = 1,y = 2,依此类推?我是否检查1&lt; 2 || 2&lt; 2(1)?
我编写了另外三个只使用'x'但没有输入10个数字的线索,这个数字是x,哪个是y。并且它不会要求用户输入x和y只有十个整数。
我会发布我所做的其他陈述的代码。
using std::cout;
using std::cin;
void ASearch(int arr[], int);
void BSearch(int array[], int size);
void CSearch(int array[], int size);
int main(int argc, const char * argv[])
{
const int SIZE = 10;
int NumArray[10];
int num = 0;
int count = 0;
while (count < 10)
{
cout << "Enter Number "<< count + 1 <<": ";
cin >> NumArray[count];
++count;
}
/*if(!isdigit(num))
{
for (int i = 0; i < 9; ++i)
{
cout << "Enter Number "<< i + 2 <<": ";
cin >> NumArray[i];
}
}else cout << "Error: numbers only";*/
for(int j = 0; j < SIZE; ++j)
{
cout << " " << NumArray[j];
}
ASearch(NumArray, SIZE);
BSearch(NumArray, SIZE);
CSearch(NumArray, SIZE);
return 0;
}
void ASearch(int arr[], int size)
{
int pos = 0;
int on = -1;
int oddTrue = 0;
int oddFalse = 0;
while (on == -1 && pos < size)
{
cout << " \nchecking: " << arr[pos];
if (arr[pos] % 2 != 0 && arr[pos] > 0)
{
cout <<"\nTrue: " <<arr[pos] << " is > 0 and odd";
++oddTrue;
}
else if(arr[pos] % 2 != 0 && arr[pos] < 0)
{
cout <<"\nFalse" << arr[pos] << "is < 0 and odd";
++oddFalse;
}
else cout << "\nNumber is even";
++pos;
}
if (oddFalse > 0) cout<< "\n>>>>>>>>>A is FALSE";else cout <<"\n>>>>>>A IS TRUE";
return;
}
void BSearch(int array[], int size)
{
int on = -1;
int pos = 0;
int count = 0;
while(on == -1 && pos < size)
{
cout << "\n checking "<<array[pos];
if(array[pos] % 2 != 0 && array[pos] > 10)
{
cout << "\nTrue: " << array[pos] <<" is > 10 & odd";
++count;
}else
{
cout <<"\nFalse";
--count;
}
++pos;
}
if(count > 1 ) cout << "\n>>>>>>>>>>B IS TRUE"; else cout <<"\n>>>>>>>>>B IS FALSE";
return;
}
void CSearch(int array[], int size)
{
int pos = 0;
int on = -1;
int TrueCount = 0;
int FalseCount = 0;
while (on == -1 && pos < size)
{
cout << "\n checking "<<array[pos];
if(array[pos] % 2 == 0 || array[pos] % 3 == 0)
{
cout << "\nTrue: " << array[pos] <<" is divisible by 2 or 3";
++TrueCount;
}
else if(array[pos] % 2 != 0 || array[pos] % 3 != 0)
{
cout <<"\nFalse";
++FalseCount;
}
++pos;
}
if (TrueCount < 10)
cout << "\n>>>>C is FALSE"; else cout << "\n>>>>>>>>>>C is TRUE";
}
答案 0 :(得分:2)
这意味着x和y都可以采用集合D中的每个值。
在伪代码中:
for i = 1:10
for j = 1:10
x = D[i]
y = D[j]
test x < y OR y < 2 * x
loop
loop